Binary operations

Step 1: Put (a=2) and (b=5) in the rule. Step 2: (2*5=2+5+2\cdot 5=17). Step 3: In such questions, do not miss the product term.

Step 1: The rule first requires the square of the first element. Step 2: (3*4=3^2+2(4)=9+8=17). Step 3: Complete powers and multiplication before addition.

Step 1: First find (5*2=2(5)-2=8). Step 2: Then (8*3=2(8)-3=13). Step 3: Always perform the operation inside the brackets first.

Step 1: First (2*3=2(2)-3=1). Step 2: Now (5*1=2(5)-1=9). Step 3: Write every intermediate value clearly to avoid option confusion.

Step 1: For identity (e), (\max(a,e)=a) must hold for every (a\in A). Step 2: (0) is the smallest element, so (\max(a,0)=a). Step 3: For a maximum operation, the smallest element is the identity.

Step 1: We need (\min(a,e)=a) for every (a\in A). Step 2: (3) is the greatest element, so (\min(a,3)=a). Step 3: For a minimum operation, the greatest element is the identity.

Step 1: For identity, write (a*e=a). Step 2: From (a+e-3=a), we get (e=3). Step 3: Cancel (a) from both sides and solve using the constant term.

Step 1: For inverse (x), (7*x=3) must hold. Step 2: (7+x-3=3) gives (x=-1). Step 3: While finding inverse, set the result equal to the identity element.

Step 1: Put (a*e=a), so (a+e+ae=a). Step 2: This gives (e(1+a)=0), and (e=0) works for every (a). Step 3: Quickly check the value in (e*a) also.

Step 1: For inverse (x), (2*x=0) must hold. Step 2: (2+x+2x=0) gives (3x=-2), so (x=-\frac{2}{3}). Step 3: Bring all terms to one side and solve a simple equation.

Step 1: Writing (a*e=a) gives (a+e+2ae=a). Step 2: (e(1+2a)=0), so (e=0) works for every (a). Step 3: An identity element must work for every (a), not just one value.

Step 1: For inverse (x), we need (1*x=0). Step 2: (1+x+2x=0) gives (3x=-1), so (x=-\frac{1}{3}). Step 3: Combine like terms and solve the equation carefully.

Step 1: The sum of two integers is an integer. Step 2: Adding (1) still gives an integer. Step 3: For closure, only check whether the result stays in the same set.

Step 1: Negative numbers are not natural numbers. Step 2: (2*5=2-5=-3), which is not a natural number. Step 3: One valid counterexample is enough to disprove closure.

Step 1: The sum of two rational numbers is rational. Step 2: Dividing a rational number by (2) also gives a rational number. Step 3: An average-based rule remains closed on rational numbers.

Step 1: (a*b=\frac{a+b}{2}) and (b*a=\frac{b+a}{2}). Step 2: Since (a+b=b+a), both are equal. Step 3: For commutativity, compare the values in both orders.

Step 1: (1*3=2), so ((1*3)*5=2*5=\frac{7}{2}). Step 2: (3*5=4), so (1*(3*5)=1*4=\frac{5}{2}). Step 3: The two values are different, so the operation is generally not associative.

Step 1: (a*0=a+0-a\cdot 0=a). Step 2: (0*a=0+a-0\cdot a=a) also holds. Step 3: The identity element leaves every element unchanged from both sides.

Step 1: For inverse (x), write (3*x=0). Step 2: (3+x-3x=0) gives (3-2x=0), so (x=\frac{3}{2}). Step 3: Handle negative signs carefully.

Step 1: (a*b=a+b-ab) and (b*a=b+a-ba). Step 2: Addition and multiplication do not change when order is reversed. Step 3: Simplify algebraic forms to check commutativity.

Step 1: (b*a=b+a+ba). Step 2: Since (b+a=a+b) and (ba=ab), both are equal. Step 3: To prove commutativity, write both forms side by side.

Step 1: (1*2=1-2+2=1). Step 2: (2*1=2-1+2=3), so they are not equal. Step 3: One unequal pair is enough to disprove commutativity.

Step 1: First (1*2=1+2+2=5). Step 2: Then (5*3=5+3+2=10). Step 3: Apply the operation twice according to the brackets.

Step 1: First (2*3=2+3+2=7). Step 2: Then (1*7=1+7+2=10). Step 3: With this rule, the brackets still give the same value here.

Step 1: ((a*b)*c=(a+b+2)+c+2=a+b+c+4). Step 2: (a*(b*c)=a+(b+c+2)+2=a+b+c+4). Step 3: If both forms are equal, associativity holds.